Question

Consider the balanced reactionbelow:2AL + 3Cl2 → 2AICI:How many grams of aluminum arerequired to produce 8.70 moles ofaluminum chloride?

Answer 1

The mass of aluminum requires to produce 8.7 moles of aluminum chloride is 234.726 grams

Explanation:

Given information

The number of moles of aluminum chloride = 8.7 moles

Let x represent the mass in grams of aluminum

The next process is to write the balanced chemical equation of the reaction

`2Al_((s))+3Cl_((2)(g))\text{ }\rightarrow2AlCl_3`

From the balanced equation of the reaction above, you will see that 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.

The next thing is to find the number of moles using a stoichiometry ratio.

Let x represents the number of moles of aluminum.

Recall that,

2 moles of aluminum react to give 2 moles of aluminum chloride

x moles of aluminum will give 8.7 moles of aluminum chloride

Mathematically,

`\begin{gathered} 2\text{ }\rightarrow\text{ 2} \\ x\text{ }\rightarrow\text{ 8.7} \\ \text{Cross multiply} \\ 2\cdot\text{ 8.7 = 2 }\cdot\text{ x} \\ 17.4\text{ = 2x} \\ \text{Divide both sides by 2} \\ (17.4)/(2)\text{ = }(2x)/(2) \\ x\text{ = 8.7 moles} \end{gathered}`

Since x represents the number of moles of aluminum, therefore, the number of moles of aluminum is 8.7 moles

The next step is to find the mass of aluminum using the below formula

`\text{mole = }\frac{\text{ reacting mass}}{\text{ molar mass}}`

From the periodic table, the molar mass of aluminum is 26.98 g/mol

`\begin{gathered} 8.7\text{ = }\frac{\text{ x}}{26.98} \\ \text{Cross multiply} \\ x\text{ = 8.7 }\cdot\text{ 26.98} \\ x\text{ = 234.726 grams} \end{gathered}`

Therefore, the mass of aluminum requires to produce 8.7 moles of aluminum chloride is 234.726 grams